Lecture (1) Introduction to Material properties And ... · Lecture (1) Introduction to Material...
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Lecture (1) Introduction to Material properties And Analysis of Reinforced Concrete Sections By Dr. Islam M. El‐Habbal
Transcript of Lecture (1) Introduction to Material properties And ... · Lecture (1) Introduction to Material...
Lecture (1) Introduction to Material properties
And Analysis of Reinforced Concrete Sections
By
Dr. Islam M. El‐Habbal
Stress‐Strain Curve of Concrete
0 0.001 0.002 0.003 0.004
Stre
ss
Strain
fcu Softening
0 0.001 0.002 0.003 0.004St
ress
Strain
fcu
Crushing Strain
Design Stress‐strain Diagram
0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035
Stre
ss
Strain
Tangent Modulus or Young’s Modulus
0.30 fcu
fcu
fcu/γc
Factor of Safety =1.50
Stress
Strain
fy
fu
Stress‐Strain Curve of Steel
εy εu
Design Stress‐Strain Curve of Steel Stress
Strain
fy/γs
fy
εy/γs εy
Factor of Safety =1.15
b
d t
As
Types of RC sections 1. Rectangular Section
Is used when slab is under Tension
d t
b
As
ts
B
2. T‐ Section
Is used when slab is under compression
1. Analysis of Rectangular Sections
2. Analysis of T‐section.
Analysis of Rectangular Singly Reinforced Sections using 1st principles
From Equilibrium:
Cu = Tu ……………………. (1)
0.67 fcu/γc *a*b = As * fy/γs get (a) check a/d≥ 0.10
c = a/0.80
Mu = Cu * yct or Mu = Tu * yct …………… (2)
Mu = 0.67 fcu/γc *a*b *(d‐a/2) or Mu = As * fy/γs *(d‐a/2) Get (Mu )
b
d t Mu
εy/γs
0.003
Strain Diagram
N.A. c
0.67 fcu/γc
Stress Diagram
0.67 fcu/γc a Cu
Tu
Idialized Stress Diagram
As
yct
Analysis of T‐Sections Singly Reinforced using 1st principles
d t
b
Mu
εy/γs
0.003
Strain Diagram
N.A. c
0.67 fcu/γc
Stress Diagram
0.67 fcu/γc a Cu
Tu
Idialized Stress Diagram
As
yct
ts
B
From Equilibrium:
Cu = Tu ……………………. (1)
0.67 fcu/γc *a*B = As * fy/γs get (a) if a> ts let a=ts check a/d ≥ 0.10 c = a/0.80
Mu = Cu * yct or Mu = Tu * yct …………… (2)
Mu = 0.67 fcu/γc *a*B *(d‐a/2) or Mu = As * fy/γs *(d‐a/2) Get (Mu )
1. Analysis of Rectangular Sections
Analysis of Rectangular Doubly Reinforced Sections using 1st principles
From Equilibrium:
Cu +Cs\= Tu
0.67 fcu/γc *a*b +As\ * fs\= As * fs ……………………. (1)
Assume that As & As\ are @ yield
0.67 fcu/γc *a*b +As\ * fy/γs = As * fy/γs get (a) check a/d≥ 0.10
c = a/0.80
b
Mu
εy/γs
0.003
Strain Diagram
N.A. c
0.67 fcu/γc
Stress Diagram
0.67 fcu/γc a Cu
Tu
Idialized Stress Diagram
As
yct
As\
d t
d’ εs\
Cs\
b
Mu
εs
0.003
Strain Diagram
N.A. c
0.67 fcu/γc
Stress Diagram
0.67 fcu/γc a Cu
Tu
Idialized Stress Diagram
As
yct
As\
d t
d’ εs\
Cs\
Check on Assumption: Since E=f/ε where E=Young’s modulus & f=stress & ε=strain So E=fy/εy where fy = yield stress & εy = yield strain & E=2000 t/cm2 = 2 x 105 MPa = 2 x 105 N/mm2 Hence get εy From symmetry of triangles: 0.003/c = εs
\ / (c‐d’) εs
\ = 0.003 * (c‐d’) / c
0.003/c = εs
/ (d‐c) εs = 0.003 * (d‐c) / c
Check of Assumption
εs\ ≥ εy/γs
εs ≥ εy/γs
εs\ < εy/γs
εs ≥ εy/γs εs
\ ≥ εy/γs
εs < εy/γs εs
\ < εy/γs
εs < εy/γs
No change in Calculations. Get Mu directly
Replace As’ fy By As’ fs ’ . get fs ’(c). Solve fs ’(c) & (1) to get C. Solve fs ’(c) and Mu(c) sim.
Replace As fy By As fs . get fs(c). Solve fs (c) & (1) to get C. Solve fs(c) and Mu(c) sim.
Replace As’ fy & As fy By As’ fs ’ & As fs . get fs ’(c), fs(c). Solve fs ’(c), fs (c) & (1) to get C. Solve fs ’(c), fs(c) and Mu(c) sim.
Get (Mu )
If εs\ ≥ εy/γs & εs ≥ εy/γs:
Mu = Cu * yct + Cs\ * (d‐d’) or Mu = Tu * yct + Cs
\ * (a/2‐d’) …………… (2)
Mu = 0.67 fcu/γc *a*b *(d‐a/2) + As\ * fy/γs * (d‐d’)
or Mu = As * fy/γs *(d‐a/2) + As\ * fy/γs * (a/2‐d’)
b
Mu
εs
0.003
Strain Diagram
N.A. c
0.67 fcu/γc
Stress Diagram
0.67 fcu/γc a Cu
Tu
Idialized Stress Diagram
As
yct
As\
d t
d’ εs\
Cs\
Check of Assumption
εs\ ≥ εy/γs
εs ≥ εy/γs
εs\ < εy/γs
εs ≥ εy/γs εs
\ ≥ εy/γs
εs < εy/γs εs
\ < εy/γs
εs < εy/γs
If εs\ < εy/γs & εs ≥ εy/γs:
fs\ = εs\ * E fs\ = 0.003 * E * (c‐d’) / c ………….. (2)
Solve (1) & (2) together to Get C C=a/0.8 check a/d ≥ 0.10 get fs\ Mu = Cu * yct + Cs
\ * (d‐d’) or Mu = Tu * yct + Cs
\ * (a/2‐d’) …………… (3)
Mu = 0.67 fcu/γc *a*b *(d‐a/2) + As\ * fs\ * (d‐d’)
or Mu = As * fy/γs *(d‐a/2) + As\ * fs\ * (a/2‐d’)
Solve (2) & (3) together and Get (Mu )
b
Mu
εs
0.003
Strain Diagram
N.A. c
0.67 fcu/γc
Stress Diagram
0.67 fcu/γc a Cu
Tu
Idialized Stress Diagram
As
yct
As\
d t
d’ εs\
Cs\
Check of Assumption
εs\ ≥ εy/γs
εs ≥ εy/γs
εs\ < εy/γs
εs ≥ εy/γs εs
\ ≥ εy/γs
εs < εy/γs εs
\ < εy/γs
εs < εy/γs
If εs\ ≥ εy/γs & εs < εy/γs:
fs = εs * E fs = 0.003 * E * (d‐c) / c ………….. (2)
Solve (1) & (2) together and Get C C=a/0.8 check a/d ≥ 0.10 get fs Mu = Cu * yct + Cs
\ * (d‐d’) or Mu = Tu * yct + Cs
\ * (a/2‐d’) …………… (3)
Mu = 0.67 fcu/γc *a*b *(d‐a/2) + As\ * fy/γs * (d‐d’)
or Mu = As * fs *(d‐a/2) + As\ * fy/γs * (a/2‐d’)
Solve (2) & (3) together and Get (Mu )
b
Mu
εs
0.003
Strain Diagram
N.A. c
0.67 fcu/γc
Stress Diagram
0.67 fcu/γc a Cu
Tu
Idialized Stress Diagram
As
yct
As\
d t
d’ εs\
Cs\
Check of Assumption
εs\ ≥ εy/γs
εs ≥ εy/γs
εs\ < εy/γs
εs ≥ εy/γs εs
\ ≥ εy/γs
εs < εy/γs εs
\ < εy/γs
εs < εy/γs
If εs\ < εy/γs & εs < εy/γs:
fs = εs * E & fs\ = εs\ * E fs = 0.003 * E * (d‐c) / c ………….. (2)
fs\ = 0.003 * E * (c‐d’) / c ………….. (3)
Solve (1) , (2) & (3) together and Get C C=a/0.8 check a/d ≥ 0.10 get fs & fs\ Mu = Cu * yct + Cs
\ * (d‐d’) or Mu = Tu * yct + Cs
\ * (a/2‐d’) …………… (4)
Mu = 0.67 fcu/γc *a*b *(d‐a/2) + As\ * fs\ * (d‐d’)
or Mu = As * fs *(d‐a/2) + As\ * fs\ * (a/2‐d’)
Solve (2) , (3) & (4) together and Get (Mu )
b
Mu
εs
0.003
Strain Diagram
N.A. c
0.67 fcu/γc
Stress Diagram
0.67 fcu/γc a Cu
Tu
Idialized Stress Diagram
As
yct
As\
d t
d’ εs\
Cs\
Check of Assumption
εs\ ≥ εy/γs
εs ≥ εy/γs
εs\ < εy/γs
εs ≥ εy/γs εs
\ ≥ εy/γs
εs < εy/γs εs
\ < εy/γs
εs < εy/γs
